/*
写完开店再写这个题目顿时神清气爽， 腰也不疼了， 眼也不花了
首先考虑将询问拆开， 就是查询一些到根的链和点k的关系
根据我们开店的结论， 一个点集到一个定点的距离和可以分三部分算
那么就很简单了吧QAQ， 在树上可持久化弄一下
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define mmp make_pair
#define M 200010
using namespace std;
int read()
{
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
int type, n, q, sz[M], son[M], fa[M], top[M], dfn[M], ver[M], dft, cnt, p[M], deep[M];
ll dis[M], sum[M], w[M], ans;
vector<pair<int,int> > to[M];
int lc[20000100], rc[20000100], v[20000100], rt[M];
ll t[20000100];
void dfs(int now, int f)
{
deep[now] = deep[f] + 1;
sz[now] = 1;
fa[now] = f;
for(int i = 0; i < to[now].size(); i++)
{
int vj = to[now][i].first, v = to[now][i].second;
if(vj == f) continue;
dis[vj] = dis[now] + v;
ver[vj] = v;
// deep[vj] = deep[now] + 1;
dfs(vj, now);
if(sz[vj] > sz[son[now]]) son[now] = vj;
sz[now] += sz[vj];
}
}
void dfs(int now)
{
dfn[now] = ++cnt;
w[cnt] = ver[now];
if(son[now])
{
top[son[now]] = top[now];
dfs(son[now]);
}
for(int i = 0; i < to[now].size(); i++)
{
int vj = to[now][i].first;
if(vj == fa[now] || vj == son[now]) continue;
top[vj] = vj;
dfs(vj);
}
}
void modify(int last, int &now, int l, int r, int ln, int rn)
{
now = ++cnt;
lc[now] = lc[last], rc[now] = rc[last], t[now] = t[last], v[now] = v[last];
if(l == ln && r == rn)
{
v[now]++;
return;
}
t[now] += w[rn] - w[ln - 1];
int mid = (l + r) >> 1;
if(ln > mid) modify(rc[last], rc[now], mid + 1, r, ln, rn);
else if(rn <= mid) modify(lc[last], lc[now], l, mid, ln, rn);
else modify(lc[last], lc[now], l, mid, ln, mid), modify(rc[last], rc[now], mid + 1, r, mid + 1, rn);
}
ll query(int now, int l, int r, int ln, int rn)
{
ll ans = 1ll * (w[rn] - w[ln - 1]) * v[now];
if(l == ln && r <= rn) return ans + t[now];
int mid = (l + r) >> 1;
if(rn <= mid) return ans + query(lc[now], l, mid, ln, rn);
else if(ln > mid) return ans + query(rc[now], mid + 1, r, ln, rn);
else return ans + query(lc[now], l, mid, ln, mid) + query(rc[now], mid + 1, r, mid + 1, rn);
}
void Modify(int &now, int x)
{
for(; top[x] != 1; x = fa[top[x]]) modify(now, now, 1, n, dfn[top[x]], dfn[x]);
modify(now, now, 1, n, dfn[1], dfn[x]);
}
void work(int now)
{
rt[now] = rt[fa[now]];
Modify(rt[now], p[now]);
sum[now] = dis[p[now]] + sum[fa[now]];
for(int i = 0; i < to[now].size(); i ++)
{
int vj = to[now][i].first;
if(vj == fa[now]) continue;
work(vj);
}
}
ll Query(int x, int k)
{
if(k == 0) return 0;
// cout << k << " ";
ll ans = sum[x];
ans += dis[k] * deep[x];
for(; top[k] != 1; k = fa[top[k]]) ans -= 2ll * query(rt[x], 1, n, dfn[top[k]], dfn[k]);
ans -= 2ll *query(rt[x], 1, n, dfn[1], dfn[k]);
// cout << x << " " << ans << "!!!!!!!\n";
return ans;
}
int LCA(int a, int b)
{
while(top[a] != top[b])
{
if(deep[top[a]] < deep[top[b]]) swap(a, b);
a = fa[top[a]];
}
if(deep[a] > deep[b]) swap(a, b);
return a;
}
int main()
{
type = read();
n = read(), q = read();
for(int i = 1; i < n; i++)
{
int vi = read(), vj = read(), v = read();
to[vi].push_back(mmp(vj, v));
to[vj].push_back(mmp(vi, v));
}
for(int i = 1; i <= n; i++) p[i] = read();
dfs(1, 0);
top[1] = 1;
dfs(1);
for(int i = 1; i <= n; i++) w[i] += w[i - 1];
work(1);
while(q--)
{
ll vi = read() ^ ans, vj = read() ^ ans, k = read() ^ ans;
int l = LCA(vi, vj);
ans = Query(vi, k) + Query(vj, k) - Query(l, k) - Query(fa[l], k);
cout << ans << "\n";
ans *= type;
}
return 0;
}
转载于:https://www.cnblogs.com/luoyibujue/p/10620465.html
